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Determine Units

  • Thread starter MattRSK
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  • #1
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Homework Statement


A quantity, d, depends on the angular frequency in the following manner:

d=ω C √(l/g) sin(ω K √(l/g))

Determine the units of C and K

Homework Equations


d=m
ω= sec-1
l=m
g=m/s2

I substituted this into the equation

m=s^(-1) C √(m/(m/s^2 )) sin(s^(-1) K √(m/(m/s^2 )) )

The Attempt at a Solution



I believe that when you take the sin of the brackets all the units in that bracket would have to become unit less. Therefore K would be a unit less constant?

And by cancellation C would have to be Meters for d to be meters?

Am i right in treating the equation as two separate parts?
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


A quantity, d, depends on the angular frequency in the following manner:

d=ω C √(l/g) sin(ω K √(l/g))

Determine the units of C and K

Homework Equations


d=m
ω= sec-1
l=m
g=m/s2

I substituted this into the equation

m=s^(-1) C √(m/(m/s^2 )) sin(s^(-1) K √(m/(m/s^2 )) )

The Attempt at a Solution



I believe that when you take the sin of the brackets all the units in that bracket would have to become unit less. Therefore K would be a unit less constant?
Well, let's see then. Inside the square root, (m/(m/s^2)= s^2 so taking the square root gives s and that is multiplied by s^(-1). Yes, K must be "unitless".

And by cancellation C would have to be Meters for d to be meters?
The the values outside the sin are the same, so if d is to be in m, c must be in m.

Am i right in treating the equation as two separate parts?
Yes, generally speaking, the argument of any function must be unitless and so can be treated separately. Looks like you have this completely right!
 
  • #3
20
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Hey thanks very much for that!
 

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